Download - Solution Guide, differential equation
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Lecture Notes in Mathematics
Arkansas Tech UniversityDepartment of Mathematics
A Second Course in ElementaryDifferential Equations
Solution Guide
Marcel B. FinancAll Rights Reserved
2015 Edition
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Contents
1 nth Order Ordinary Differential Equations 51.1 Calculus of Matrix-Valued Functions of a Real Variable . . . . 51.2 Uniform Convergence and Weierstrass M-Test . . . . . . . . . 131.3 nth Order Linear Differential Equations: Exsitence and Unique-
ness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 The General Solution of nth Order Linear Homogeneous Equa-
tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.5 Fundamental Sets and Linear Independence . . . . . . . . . . 321.6 nth Order Homogeneous Linear Equations with Constant Co-
efficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.7 Review of Power Series . . . . . . . . . . . . . . . . . . . . . . 411.8 Series Solutions of Linear Homogeneous Differential Equations 47
3
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4 CONTENTS
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Chapter 1
nth Order Ordinary DifferentialEquations
1.1 Calculus of Matrix-Valued Functions of a
Real Variable
Problem 1.1.1Consider the following matrices
A(t) =
[t 1 t2
2 2t+ 1
], B(t) =
[t 10 t+ 2
], c(t) =
[t+ 11
](a) Find 2A(t) - 3tB(t).(b) Find A(t)B(t) - B(t)A(t).(c) Find A(t)c(t).(d) Find det(B(t)A(t)).
Solution.(a) We have
2A(t) 3tB(t) = 2[t 1 t2
2 2t+ 1
] 3t
[t 10 t+ 2
]=
[2t 2 2t2
4 4t+ 2
][
3t2 3t0 3t2 + 6t
]=
[2t 2 3t2 2t2 + 3t
4 2 2t 3t2]
5
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6CHAPTER 1. NTH ORDER ORDINARY DIFFERENTIAL EQUATIONS
(b) We have
A(t)B(t)B(t)A(t) =[t 1 t2
2 2t+ 1
] [t 10 t+ 2
][t 10 t+ 2
] [t 1 t2
2 2t+ 1
]=
[t2 t t3 + 2t2 t+ 1
2t 2t2 + 5t
][t2 t 2 t3 2t 1
2t+ 4 2t2 + 5t+ 2
]=
[2 2t2 + t+ 24 2
](c) We have
A(t)c(t) =
[t 1 t2
2 2t+ 1
] [t+ 11
]=
[(t 1)(t+ 1) + t2(1)2(t+ 1) + (2t+ 1)(1)
]=
[11
](d) We have
det(B(t)A(t)) = det
([t 10 t+ 2
] [t 1 t2
2 2t+ 1
])= det
([t2 t 2 t3 2t 1
2t+ 4 2t2 + 5t+ 2
]= (t3 + 3t2 + 2t)
Problem 1.1.2Determine all values t such that A(t) is invertible and, for those t-values,find A1(t).
A(t) =
[t+ 1 tt t+ 1
].
Solution.We have det(A(t)) = 2t + 1 so that A is invertible for all t 6= 1
2. In this
case,
A1(t) =1
2t+ 1
[t+ 1 tt t+ 1
]Problem 1.1.3Determine all values t such that A(t) is invertible and, for those t-values,find A1(t).
A(t) =
[sin t cos tsin t cos t
].
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1.1. CALCULUS OFMATRIX-VALUED FUNCTIONS OF A REAL VARIABLE7
Solution.We have det(A(t)) = 2 sin t cos 2 = sin 2t so that A is invertible for all t 6= n
2
where n is an integer. In this case,
A1(t) =1
sin 2t
[cos t cos t sin t sin t
]Problem 1.1.4Find
limt0
[sin tt
t cos t 3t+1
e3t sec t 2tt21
].
Solution.
limt0
[sin tt
t cos t 3t+1
e3t sec t 2tt21
]=
[1 0 31 1 0
]Problem 1.1.5Find
limt0
[tet tan tt2 2 esin t
].
Solution.
limt0
[tet tan tt2 2 esin t
]=
[0 02 1
]Problem 1.1.6Find A(t) and A(t) if
A(t) =
[sin t 3tt2 + 2 5
].
Solution.
A(t) =
[cos t 32t 0
]A(t) =
[ sin t 0
2 0
]
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8CHAPTER 1. NTH ORDER ORDINARY DIFFERENTIAL EQUATIONS
Problem 1.1.7Express the system
y1 = t2y1 + 3y2 + sec t
y2 = (sin t)y1 + ty2 5
in the matrix form
y(t) = A(t)y(t) + g(t).
Solution.
y(t) =
[y1(t)y2(t)
], A(t) =
[t2 3
sin t t
], g(t) =
[sec t5
]Problem 1.1.8Determine A(t) where
A(t) =
[2t 1
cos t 3t2
], A(0) =
[2 51 2
].
Solution.Integrating componentwise we find
A(t) =
[t2 + c11 t+ c12
sin t+ c21 t3 + c22
]Since
A(0) =
[2 51 2
]=
[c11 c12c21 c22
]by equating componentwise we find c11 = 2, c12 = 5, c21 = 1, and c22 = 2.Hence,
A(t) =
[t2 + 2 t+ 5
sin t+ 1 t3 +2
]Problem 1.1.9Determine A(t) where
A(t) =
[1 t0 0
], A(0) =
[1 12 1
], A(0) =
[1 22 3
].
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1.1. CALCULUS OFMATRIX-VALUED FUNCTIONS OF A REAL VARIABLE9
Solution.Integrating componentwise we find
A(t) =
[t+ c11
t2
2+ c12
c21 c22
]Integrating again we find
A(t) =
[t2
2+ c11t+ d11
t3
6+ c12t+ d12
c21t+ d21 c22t+ d22
]But
A(0) =
[1 12 1
]=
[d11 d12d21 d22
]so by equating componentwise we find d11 = 1, d12 = 1, d21 = 2, andd22 = 1. Thus,
A(t) =
[t2
2+ c11t+ 1
t3
6+ c12t+ 1
c21t+2 c22t+ 1
]Since
A(0) =
[1 22 3
]=
[c11 c12c21 c22
]we find c11 = 1, c12 = 2, c21 = 2, c22 = 3. Hence,
A(t) =
[t2
2 t+ 1 t3
6+ 2t+ 1
4 3t+ 1
]Problem 1.1.10Calculate A(t) =
t0
B(s)ds where
B(s) =
[es 6s
cos 2s sin 2s
].
Solution.Integrating componentwise we find
A(t) =
[ t0esds
t0
6sds t0
cos 2sds t0
sin 2sds
]=
[et 1 3t2sin 2t2
1cos 2t2
]
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10CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
Problem 1.1.11Construct a 2 2 nonconstant matrix function A(t) such that A2(t) is aconstant matrix.
Solution.Let
A(t) =
[0 t0 0
]Then
A2(t) =
[0 t0 0
] [0 t0 0
]=
[0 00 0
]Problem 1.1.12(a) Construct a 2 2 differentiable matrix function A(t) such that
d
dtA2(t) 6= 2A d
dtA(t).
That is, the power rule is not true for matrix functions.(b) What is the correct formula relating A2(t) to A(t) and A(t)?
Solution.(a) Let
A(t) =
[1 tt2 0
]. Then
A2(t) =
[1 + t3 tt2 t3
]so that
d
dtA2(t) =
[3t2 12t 3t2
].
On the other hand,
2A(t)d
dtA(t) = 2
[1 tt2 0
] [1 02t 0
]=
[2 + 4t2 0
2t2 0
](b) The correct formula is d
dtA2(t) = A(t)A(t) + A(t)A(t)
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1.1. CALCULUS OFMATRIX-VALUED FUNCTIONS OF A REAL VARIABLE11
Problem 1.1.13Transform the following third-order equation
y 3ty + (sin 2t)y = 7et
into a first order system of the form
x(t) = Ax(t) + b(t).
Solution.Let x1 = y, x2 = y
, x3 = y. Then by letting
x =
x1x2x3
, A = 0 1 00 0 1 sin 2t 3t 0
, b = 00
7et
the differential equation can be presented by the first order system
x(t) = Ax(t) + b(t)
Problem 1.1.14By introducing new variables x1 and x2, write y
2y+ 1 = t as a system oftwo first order linear equations of the form x + Ax = b.
Solution.By letting x1 = y and x2 = y
we have
x =
[x1x2
], A =
[0 12 0
], b =
[0
t 1
]Problem 1.1.15Write the differential equation y + 4y + 4y = 0 as a first order system ofthe form x(t) + A(t)x(t) = b(t).
Solution.By letting x1 = y and x2 = y
we have
x =
[x1x2
], A =
[0 14 4
], b =
[00
]
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12CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
Problem 1.1.16Write the differential equation y+ ky+ (t 1)y = 0 as a first order systemof the form x(t) + A(t)x(t) = b(t)..
Solution.By letting x1 = y and x2 = y
we have
x =
[x1x2
], A =
[0 1
t 1 k
], b =
[00
]Problem 1.1.17Change the following second-order equation to a first-order system.
y 5y + ty = 3t2, y(0) = 0, y(0) = 1Solution.If we write the problem in the matrix form
x + Ax = b, x(0) = y0
then
A =
[0 1t 5
]x =
[x1x2
]=
[yy
], b =
[0
3t2
], y0 =
[01
]Problem 1.1.18Consider the following system of first-order linear equations.
x =
[3 21 1
]x.
Find the second-order linear differential equation that x satisfies.
Solution.The system is
x1 = 3x1 + 2x2x2 = x1 x2
It follows that
x1 + 2x2 = 5x1 or x1 =
x1+2x2
5
so we let w = x1+2x25
so that w = x1. Thus, x1 = 3x1+2x2 = 3x1+(5wx1) =
2x1 + 5w. Hence, x1 = 2x
1 + 5w
= 2x1 + 5x1 or x1 2x1 5x1 = 0.
Likewise, x22x25x2 = 0. Hence, x1 and x2 satisfy the differential equationy 2y 5y = 0
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1.2. UNIFORM CONVERGENCE AND WEIERSTRASS M-TEST 13
1.2 Uniform Convergence and Weierstrass M-
Test
Problem 1.2.1Define fn : [0, 1] R by fn(x) = xn. Define f : [0, 1] R by
f(x) =
{0 if 0 x < 11 if x = 1
(a) Show that the sequence {fn}n=1 converges pointwise to f.(b) Show that the sequence {fn}n=1 does not converge uniformly to f. Hint:Suppose otherwise. Let = 0.5 and get a contradiction by using a point(0.5)
1N < x < 1.
Solution.(a) For all 0 x < 1 we have limn fn(x) = limn xn = 0. Also,limn fn(1) = 1. Hence, the sequence {fn}n=1 converges pointwise to f.(b) Suppose the contrary. Let = 1
2. Then there exists a positive integer N
such that for all n N we have
|fn(x) f(x)| 0 be given. Let N be a positive integer such that N > 1
. Then for
n N x xnn x = |x|nn < 1n 1N < .
Thus, the given sequence converges uniformly (and pointwise) to the functionf(x) = x
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16CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
Problem 1.2.7Let fn(x) =
xn
1+xnfor x [0, 2].
(a) Find the pointwise limit f(x) = limn fn(x) on [0, 2].(b) Does fn f uniformly on [0, 2]?
Solution.(a) The pointwise limit is
f(x) =
0 if 0 x < 112
if x = 11 if 1 < x 2.
(b) The convergence cannot be uniform because if it were, f would have tobe continuous since each fn is continuous on [0, 2]
Problem 1.2.8For each n N define fn : R R by fn(x) = n+cosx2n+sin2 x .(a) Show that fn 12 uniformly.(b) Find limn
72fn(x)dx.
Solution.(a) Let > 0 be given. Note that
|fn(x)1
2| =
2 cosx sin2 x2(2n+ sin2 x) 34n.
Since limn34n
= 0 we can find a positive integer N such that if n Nthen 3
4n< . Thus, for n N and all x R we have
|fn(x)1
2| 3
4n< .
This shows that fn 12 uniformly on R and also on [2, 7].(b) We have
limn
72
fn(x)dx =
72
limn
fn(x)dx =
72
1
2dx =
5
2
Problem 1.2.9Show that the sequence defined by fn(x) = (cosx)
n does not converge uni-formly on [
2, 2].
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1.2. UNIFORM CONVERGENCE AND WEIERSTRASS M-TEST 17
Solution.We have proved in Problem 1.2.5 that this sequence converges pointwise tothe discontinuous function
f(x) =
{0 if
2 x < 0 and 0 < x
2
1 if x = 0.
Therefore, uniform convergence cannot occur for this given sequence
Problem 1.2.10Let {fn}n=1 be a sequence of functions such that
sup{|fn(x)| : 2 x 5} 2n
1 + 4n.
(a) Show that this sequence converges uniformly to a function f to be found.
(b) What is the value of the limit limn 52fn(x)dx?
Solution.(a) Using the squeeze rule we find
limn
sup{|fn(x)| : 2 x 5} = 0.
Let > 0 be given. Then there is a positive integer N such that for n N,we have
sup{|fn(x)| : 2 x 5} < .Thus,
|fn(x)| sup{|fn(x)| : 2 x 5} < for all x [2, 5]. Thus, {fn}n=1 converges uniformly to the zero function.(b) We have
limn
52
fn(x)dx =
52
limn
fn(x)dx =
52
0dx = 0
Problem 1.2.11Show that the series
n=1
12n
cos (3nx) converges uniformly on R.
Solution.We have 12n cos (3nx)
12n = Mn.Since
n=1
12n
= 12
11 1
2
= 1, by the Weierstrass M-test the given series con-
verges uniformly on R
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18CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
Problem 1.2.12Show that the series
n=1
nn4+x4
converges uniformly on R.
Solution.Since n4 + x4 n4 for all x R, we have nn4 + x4
nn4 = 1n3 = Mn.Since
n=1
1n3
is a pseries with p = 3 > 1 so it is convergent. By theWeierstrass M-test the given series converges uniformly on R
Problem 1.2.13Show that the series
n=1
1n2+x2
converges uniformly on R.
Solution.Since n2 + x2 n2 for all x R, we have 1n2 + x2
1n2 = Mn.Since
n=1
1n2
is a pseries with p = 2 > 1 so it is convergent. By theWeierstrass M-test the given series converges uniformly on R
Problem 1.2.14Suppose that
n=1 |an|
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1.3. NTH ORDER LINEAR DIFFERENTIAL EQUATIONS: EXSITENCE ANDUNIQUENESS19
1.3 nth Order Linear Differential Equations:
Exsitence and Uniqueness
Problem 1.3.1Convert the initial value problem
y 1t2 9
y + ln (t+ 1)y + (cos t)y = 0, y(0) = 1, y(0) = 3, y(0) = 0
into the matrix form
x(t) = A(t)x(t) + b(t), x(t0) = x0.
Solution.Letting x1 = y, x2 = y
, and x3 = y we find
A(t) =
0 1 00 0 1 cos t ln (t+ 1) 1
t29
x(t) =
x1x2x3
, b(t) =00
0
, x(0) =13
0
Problem 1.3.2Convert the initial value problem
y +1
t+ 1y + (tan t)y = 0, y(0) = 0, y(0) = 1, y(0) = 2
into the matrix form
x(t) = A(t)x(t) + b(t), x(t0) = x0.
Solution.Letting x1 = y, x2 = y
, and x3 = y we find
A(t) =
0 1 00 0 1 tan t 1
t+10
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20CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
x(t) =
x1x2x3
, b(t) =00
0
, x(0) =01
2
Problem 1.3.3Convert the initial value problem
y 1t2 + 9
y+ln (t2 + 1)y+(cos t)y = t3+t5, y(0) = 1, y(0) = 3, y(0) = 0
into the matrix form
x(t) = A(t)x(t) + b(t), x(t0) = x0.
Solution.Letting x1 = y, x2 = y
, and x3 = y we find
A(t) =
0 1 00 0 1 cos t ln (t2 + 1) 1
t2+9
x(t) =
x1x2x3
, b(t) = 00t3 + t 5
, x(0) =13
0
Problem 1.3.4Transform the following third-order equation
y 3ty + (sin 2t)y = 7et
into a first order system of the form
x(t) = Ax(t) + b(t)
Solution.Let x1 = y, x2 = y
, x3 = y. Then by letting then the differential equation
can be presented by the first order system
x(t) = Ax(t) + b(t)
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1.3. NTH ORDER LINEAR DIFFERENTIAL EQUATIONS: EXSITENCE ANDUNIQUENESS21
Problem 1.3.5If y(t) = 14
33e4t + 13
15e2t 16
55e7t is the solution to the initial value problem
y 5y 22y + 56y = 0, y(0) = 1, y(0) = 2, y(0) = 4
then what is the solution to the corresponding matrix system
x(t) = A(t)x(t) + b(t), x(t0) = x0?
Solution.The solution to the matrix system is
x(t) =
yyy
= 1433e4t + 1315e2t 1655e7t56
33e4t + 26
15e2t 112
55e7t
22433e4t + 52
15e2t 784
55e7t
Problem 1.3.6Suppose that the vector
x(t) =
et + et + cos t+ sin tet et sin t+ cos tet + et cos t sin tet et + sin t cos t
is a solution to the matrix system
x(t) =
0 1 0 00 0 1 00 0 0 11 0 0 0
x.Find the ordinary differential equation with solution y(t) = et + et + cos t+sin t.
Solution.We have y(4) = x
(4)1 = x
4 = (e
tet+sin tcos t) = et+et+cos t+sin t = y.Thus, y satisfies the equation y(4) y = 0
Problem 1.3.7Suppose that the vector
x(t) =
1 + cos (5t) + sin (5t)5 sin (5t) + 5 cos (5t)25 cos (5t) 25 sin (5t)
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22CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
is a solution to the matrix system
x(t) =
0 1 00 0 10 25 0
x.Find the ordinary differential equation with solution y(t) = 1 + cos (5t) +sin (5t).
Solution.We have y = x
(3)1 = x
3 = (25 cos (5t) 25 sin (5t)) = 125 sin (5t)
125 cos (5t) = 25y. Thus, y satisfies the equation y + 25y = 0
Problem 1.3.8Find the first two Picards iterations in Problem 1.3.1.
Solution.We have
x(t) = x(0) +
t0
A(s)x(s)ds.
Thus,
x0(t) =
130
x1(t) =
130
+ t0
0 1 00 0 1 cos s ln (s+ 1) 1
s29
130
ds=
130
+ t0
30 cos s 3 ln (s+ 1)
ds=
130
+ 3t0 sin t 3(t+ 1) ln (t+ 1) + 3t
=
1 + 3t3 sin t 3(t+ 1) ln (t+ 1) + 3t
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1.3. NTH ORDER LINEAR DIFFERENTIAL EQUATIONS: EXSITENCE ANDUNIQUENESS23
Problem 1.3.9Find the first two Picards iterations in Problem 1.3.3.
Solution.We have
x(t) = x(0) +
t0
A(s)x(s)ds.
Thus,
x0(t) =
130
x1(t) =
130
+ t0
0 1 00 0 1 cos s ln (s2 + 1) 1
s2+9
130
ds=
130
+ t0
30 cos s 3 ln (s2 + 1)
ds=
130
+ 3t0 sin t 3t ln (t2 + 1) + 6t 6 tan1 t
=
1 + 3t3 sin t 3t ln (t2 + 1) + 6t 6 tan1 t
Problem 1.3.10Prove the inquality (1.3.6) by induction on N 1.
Solution.Basis of Induction: For N = 1, we have
||x1 x0|| M(t t0) = MK11(t t0)1
1!.
Induction Hypothesis: Suppose
||xk xk1|| MKk1(t t0)k
k!.
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24CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
For k = 1, 2, , N.Induction Step: We want to show that
||xN+1 xN || MKN(t t0)N+1
(N + 1)!.
Indeed, we have
||xN+1 xN || K tt0
||xN xN1||ds
K tt0
MKN1(s t0)N
N !ds
=MKN(t t0)N+1
(N + 1)!
Problem 1.3.11Let f(t) be a continuous function such that f(t) C +K
taf(s)ds for some
constants C and K 0. Show that f(t) CeK(ta).
Solution.This follows from Gronwalls lemma with h(t) = K
Problem 1.3.12Let f(t) be a non-negative continuous function such that f(t)
taf(s)ds
for all t a. Show that f(t) 0 for all t a.
Solution.This follows from Gronwalls lemma with C = 0 and h(t) = 1
For Problems 1.3.12 - 1.3.15, use Theorem 1.3.1 to find the largest inter-val a < t < b in which a unique solution is guaranteed to exist.
Problem 1.3.13
y 1t2 9
y + ln (t+ 1)y + (cos t)y = 0, y(0) = 1, y(0) = 3, y(0) = 0
Solution.The coefficient functions are all continuous for t 6= 3, 3 and t > 1. Sincet0 = 0, the largest interval of existence is 1 < t < 3
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1.3. NTH ORDER LINEAR DIFFERENTIAL EQUATIONS: EXSITENCE ANDUNIQUENESS25
Problem 1.3.14
y +1
t+ 1y + (tan t)y = 0, y(0) = 0, y(0) = 1, y(0) = 2
Solution.The coefficient functions are all continuous for t 6= 1 and t 6= (2n + 1)
2
where n is an integer. Since t0 = 0, the largest interval of existence is1 < t <
2
Problem 1.3.15
y 1t2 + 9
y + ln (t2 + 1)y + (cos t)y = 0, y(0) = 1, y(0) = 3, y(0) = 0
Solution.The coefficient functions are all continuous for t so that the interval of exis-tence is < t 0, we must have 0 = r3 2r2 r + 2 = r2(r 2) (r 2) =(r 2)(r2 1) = (r 2)(r 1)(r + 1). Hence, r = 1, 1, 2
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26CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
1.4 The General Solution of nth Order Linear
Homogeneous Equations
In Problems 1.4.1 - 1.4.3, show that the given solutions form a fundamentalset for the differential equation by computing the Wronskian.
Problem 1.4.1
y y = 0, y1(t) = 1, y2(t) = et, y3(t) = et.
Solution.We have
W (t) =
1 et et
0 et et0 et et
= (1)
et etet et et 0 et0 et
+ et 0 et0 et = 2 6= 0
Problem 1.4.2
y(4) + y = 0, y1(t) = 1, y2(t) = t, y3(t) = cos t, y4(t) = sin t.
Solution.We have
W (t) =
1 t cos t sin t0 1 sin t cos t0 0 cos t sin t0 0 sin t cos t
= cos2 t+ sin2 t = 1 6= 0
Problem 1.4.3
t2y + ty y = 0, y1(t) = 1, y2(t) = ln t, y3(t) = t2.
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1.4. THE GENERAL SOLUTION OFNTH ORDER LINEAR HOMOGENEOUS EQUATIONS27
Solution.We have
W (t) =
1 ln t t2
0 1t
2t0 1
t22
= (1)
t1 2tt2 2 ln t 0 2t0 2
+ t2 0 t10 t2 =
=3t1 6= 0, t > 0
Use the fact that the solutions given in Problems 1.4.1 - 1.4.3 for a funda-mental set of solutions to solve the following initial value problems 1.4.4 -1.4.6.
Problem 1.4.4
y y = 0, y(0) = 3, y(0) = 3, y(0) = 1.
Solution.The general solution is y(t) = c1 + c2e
t + c3et. With the initial conditions
we havey(0) = 3 = c1 + c2 + c3 = 3y(0) = 3 = c2 c3 = 3y(0) = 1 = c2 + c3 = 1
Solving these simultaneous equations gives c1 = 2, c2 = 1 and c3 = 2 andso the unique solution is
y(t) = 2 et + 2et
Problem 1.4.5
y(4) + y = 0, y(
2) = 2 + , y(
2) = 3, y(
2) = 3, y(
2) = 1.
Solution.The general solution is y(t) = c1 + c2t + c3 cos t + c4 sin t. With the initialconditions we have
y(2) = 2 + = c1 + 2 c2 + c4 = 2 +
y(2) = 3 = c2 c3 = 3
y(2) = 3 = c4 = 3
y(2) = 1 = c3 = 1
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28CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
Solving these simultaneous equations gives c1 = ( + 1), c2 = 4, c3 = 1and c4 = 3. Hence, the unique solution is
y(t) = ( + 1) + 4t+ cos t+ 3 sin t
Problem 1.4.6
t2y + ty y = 0, , y(1) = 1, y(1) = 2, y(1) = 6.
Solution.The general solution is y(t) = c1 + c2 ln t + c3t
2. With the initial conditionswe have
y(1) = 1 = c1 + c3 = 1y(1) = 2 = c2 + 2c3 = 2y(1) = 6 = c2 + 2c3 = 6
Solving these simultaneous equations gives c1 = 2, c2 = 4 and c3 = 1 andso the unique solution is
y(t) = 2 + 4 ln t t2
Problem 1.4.7In each question below, show that the Wronskian determinant W (t) behavesas predicted by Abels Theorem. That is, for the given value of t0, show that
W (t) = W (t0)e
tt0pn1(s)ds.
(a) W (t) found in Problem 1.4.1 and t0 = 1.(b) W (t) found in Problem 1.4.2 and t0 = 1.(c) W (t) found in Problem 1.4.3 and t0 = 2.
Solution.(a) For the given differential equation pn1(t) = p2(t) = 0 so that Abelstheorem predicts W (t) = W (t0). Now, for t0 = 1 we have W (t) = W (1) =constant. From Problem 1.4.1, we found that W (t) = W (1)e
t1 0ds = 2.
(b) For the given differential equation pn1(t) = p3(t) = 0 so that Abelstheorem predict W (t) = W (t0). Now, for t0 = 1 we have W (t) = W (1) =
constant. From Problem 1.4.2, we found that W (t) = W (1)e t1 0ds = 1.
(c) For the given differential equation pn1(t) = p2(t) =1t
so that Abels
theorem predict W (t) = W (2)e t2
dss = W (2)eln (
2t) = 2
tW (2). From Problem
1.4.3, we found that W (t) = 3t
= W (2)e t2
dss where W (2) = 3
2
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1.4. THE GENERAL SOLUTION OFNTH ORDER LINEAR HOMOGENEOUS EQUATIONS29
Problem 1.4.8Determine W (t) for the differential equation y+(sin t)y+(cos t)y+2y = 0such that W (1) = 0.
Solution.Here pn1(t) = p2(t) = sin t. By Abels Theorem we have
W (t) = W (1)e t1 sin sds 0
Problem 1.4.9Determine W (t) for the differential equation t3y2y = 0 such that W (1) =3.
Solution.Here pn1(t) = p2(t) = 0. By Abels Theorem we have
W (t) = W (1)e t1 0ds = W (1) = 3
Problem 1.4.10Consider the initial value problem
y y = 0, y(0) = , y(0) = , y(0) = 4.
The general solution of the differential equation is y(t) = c1 + c2et + c3e
t.(a) For what values of and will limt y(t) = 0?(b) For what values and will the solution y(t) be bounded for t 0, i.e.,|y(t)| M for all t 0 and for some M > 0? Will any values of and produce a solution y(t) that is bounded for all real number t?
Solution.(a) Since y(0) = , c1 + c2 + c3 = . Since y
(t) = c2et c3et and y(0) = ,
c2c3 = . Also, since y(t) = c2et+c3et and y(0) = 4 we have c2+c3 = 4.Solving these equations for c1, c2, and c3 we find c1 = 4, c2 = /2 + 2and c3 = /2 + 2. Thus,
y(t) = 4 + (/2 + 2)et + (/2 + 2)et.
If limt y(t) = 0, then we must have 4 = 0 and 2 + 2 = 0. In this case, = 4 and = 4. Thus,
y(t) = 4et.
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30CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS
(b) In the expression of y(t) we know that et is bounded for t 0 (et 1for t 0) whereas et is unbounded for t 0. Thus, for y(t) to be boundedwe must choose
2+ 2 = 0 or = 4. The number can be any number.
Now, for the solution y(t) to be bounded on < t < we must havesimultaneously /2+2 = 0 and /2+2 = 0. But there is no that satisfiesthese two equations at the same time. Hence, y(t) is always unbounded forany choice of and
Problem 1.4.11Consider the differential equation y + p2(t)y
+ p1(t)y = 0 on the interval
1 < t < 1. Suppose it is known that the coefficient functions p2(t) and p1(t)are both continuous on 1 < t < 1. Is it possible that y(t) = c1 + c2t2 + c3t4is the general solution for some functions p1(t) and p2(t) continuous on 1